NCERT Solutions for Class 8 Maths Chapter 6 Cubes And Cube Roots Ex 6.2

Exercise 6.2

1. Find the cube root of each of the following numbers by prime factorisation method

i. $64$

Ans: Expand $64$ in factors of prime numbers.

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} $

Take cube root on both sides of equation.

$ \because 64 = {2^3} \times {2^3} $

  $\therefore \sqrt[3]{{64}} = 2 \times 2 = 4 $

The cube root of $64$ is $4.$

ii. $512$

Ans: Expand $512$ in factors of prime numbers.

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} \times {2^3} $

Take cube root on both sides of equation.

$\because 512 = {2^3} \times {2^3} \times {2^3} $

$\therefore \sqrt[3]{{512}} = 2 \times 2 \times 2 = 8 $

The cube root of $512$ is $8.$

iii. $10648$

Ans: Expand $10648$ in factors of prime numbers.

$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 $

$= {2^3} \times {11^3} $

Take cube root on both sides of equation.

$\because 10648 = {2^3} \times {11^3} $

$\therefore \sqrt[3]{{10648}} = 2 \times 11 = 22 $

The cube root of $10648$ is $22.$

iv. $27000$

Ans: Expand $27000$ in factors of prime numbers.

$ 27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {2^3} \times {3^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 27000 = {2^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{27000}} = 2 \times 3 \times 5 = 30 $

The cube root of $27000$ is $30.$

v. $15625$

Ans: Expand $15625$ in factors of prime numbers.

$  15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 $

$ = {5^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 15625 = {5^3} \times {5^3} $

$\therefore \sqrt[3]{{15625}} = 5 \times 5 = 25 $

The cube root of $15625$ is $25.$

vi. $13824$

Ans: Expand $13824$ in factors of prime numbers.

$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take cube root on both sides of equation.

$  \because 13284 = {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{13284}} = 2 \times 2 \times 2 \times 3 = 24 $

The cube root of $13824$ is $24$

vii. $110592$

Ans: Expand $110592$ in factors of prime numbers.

$115092 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take cube root on both sides of equation.

$\because 110592 = {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{110592}} = 2 \times 2 \times 2 \times 2 \times 3 = 48 $

The cube root of $110592$ is $48.$

viii. $46656$

Ans: Expand $46656$ in factors of prime numbers.

$46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {3^3} \times {3^3} $

Take cube root on both sides of equation.

$\because 46656 = {2^3} \times {2^3} \times {3^3} \times {3^3} $

$\therefore \sqrt[3]{{46656}} = 2 \times 2 \times 3 \times 3 = 36 $

The cube root of $46656$ is $36.$

ix. $175616$

Ans: Expand $175616$ in factors of prime numbers.

$175616 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 $

$= {2^3} \times {2^3} \times {2^3} \times {7^3} $

Take cube root on both sides of equation.

$\because 175616 = {2^3} \times {2^3} \times {2^3} \times {7^3} $

$\therefore \sqrt[3]{{175616}} = 2 \times 2 \times 2 \times 7 = 56 $

The cube root of $175616$ is $56.$

x. $91125$

Ans: Expand $91125$ in factors of prime numbers.

$91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {3^3} \times {3^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 91125 = {3^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{91125}} = 3 \times 3 \times 5 = 45 $

The cube root of $91125$  is $45.$

2. State true or false.

i) Cube of any odd number is even. 

Ans: False. As multiplying an odd number three times will yield an odd number.

For example, the cube of $5$ which is an odd number is \[25\], which is also an odd number.

ii) A perfect cube does not end with two zeroes. 

Ans: True. Perfect cube will always terminate with multiple of $3$ numbers of zeroes.

For example, the cube of \[100\] is $1000000$ and there are $6$ zeros at the end of it.

iii) If square of a number ends with \[5\], then its cube ends with \[25\]. 

Ans: False, it is not always certain that if the square of a number ends with \[5\], then its cube will end with \[25\].

For examples, square of $55$ ends with 5, $3025$ but its cube,$166375$ does not end with \[25\].

iv) There is no perfect cube which ends with \[8\]. 

Ans: False, all the numbers having \[2\] at its unit digit place will have \[8\] in end as cube.

v) The cube of a two digit number may be a three digit number. 

Ans: False, as cube of even smallest two digit number, $10$ is a four digit number,$1000$.

vi) The cube of a two digit number may have seven or more digits. 

Ans: False, as cube of even largest two digit number, $99$ is a six digit number, $970299$

vii) The cube of a single digit number may be a single digit number. 

Ans: True, as a cube of first two natural numbers, $1$ and $2$ are $1$ and $8$ respectively.

Conclusion

Class 8 Maths Ch 6 Ex 6.2 focuses on cubes and cube roots and calculating cube root, providing essential practice for understanding these concepts. It is important to understand the properties of cubes and cube roots, as well as how to calculate them efficiently. Pay special attention to the methods for finding cubes and cube roots, as these skills are foundational for more advanced mathematical topics. This exercise is crucial for developing strong problem-solving abilities and a deeper understanding of numerical relationships, which will benefit students in their future studies.

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Chapter-Specific NCERT Solutions for Class 8 Maths

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